If 2.00 grams of CO2 were produced when 4.00 grams of C5H12 were combusted in excessoxygen, what is the percent yield of the combustion reaction?

The % yield of the reaction is 16.4 %

Explanation:

Step 1: Data given

Mass of CO2 = 2.00 grams

Mass of C5H12 = 4.00 grams

O2 is in excess

Molar mass of CO2 = 44.01 g/mol

Molar mass of C5H12 = 72.15 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

C5H12 + 8O2 → 5CO2 + 6H2O

Step 3: Calculate moles C5H12

Moles C5H12 = mass C5H12 / molar mass C5H12

Moles C5H12 = 4.00 grams / 72.15 g/mol

Moles C5H12 = 0.0554 moles

Step 4: Calculate moles of CO2

For 1 mol pentane, we need 8 moles of O2 to produce 5 moles CO2 and 6 moles H2O

For 0.0554 moles C5H12 we'll have 5*0.0554 = 0.277 moles CO2

Step 5: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.277 moles ¨44.01 g/mol

Mass CO2 = 12.2 grams CO2

Step 6: Calculate % yield

% yield = (actual yield / theoretical yield) * 100%

% yield = (2.00 grams / 12.2 grams) * 100%

% yield = 16.4 %

The % yield of the reaction is 16.4 %